Unique Binary Search Trees

来源：互联网发布：php.ini file uploads 编辑：程序博客网时间：2024/04/30 05:45
问题来源：https://leetcode.com/problems/unique-binary-search-trees/
/** *  * <p> * ClassName UniqueBinarySearchTrees * </p> * <p> * Description Given n, how many structurally unique BST's (binary search trees二叉查找树) that store values 1...n? For example, Given * n = 3, there are a total of 5 unique BST's.<br/> *   1         3     3      2      1<br/> *    \       /     /      / \      \<br/> *     3     2     1      1   3      2<br/> *    /     /       \                 \<br/> *   2     1         2                 3<br/> * </p> *  * @author TKPad wangx89@126.com *         <p> *         Date 2015年3月20日 下午2:11:23 *         </p> * @version V1.0.0 * * *//** * 卡特兰数又称卡塔兰数，英文名Catalan number，是组合数学中一个常出现在各种计数问题中出现的数列。由以比利时的数学家欧仁·查理·卡塔兰 (1814–1894)命名，其前几项为 : 1, 2, 5, 14, 42, 132, 429, 1430, * 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, * 91482563640, 343059613650, 1289904147324, 4861946401452 *//** * 令h(0)=1,h(1)=1，catalan数满足递推式[1] ： h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2) 例如：h(2)=h(0)*h(1)+h(1)*h(0)=1*1+1*1=2 * h(3)=h(0)*h(2)+h(1)*h(1)+h(2)*h(0)=1*2+1*1+2*1=5 另类递推式[2] ： h(n)=h(n-1)*(4*n-2)/(n+1); 递推关系的解为： h(n)=C(2n,n)/(n+1) * (n=0,1,2,...) 递推关系的另类解为： h(n)=c(2n,n)-c(2n,n+1)(n=0,1,2,...) */public class UniqueBinarySearchTrees {    // Time Limit Exceeded    // public int numTrees(int n) {    // if (0 == n || 1 == n) {    // return 1;    // } else {    // int sum = 0;    // for (int i = 0; i <= n - 1; i++) {    // sum += numTrees(i) * numTrees(n - 1 - i);    // }    // return sum;    // }    // }    public int numTrees(int n) {        if (n <= 1) {            return 1;        }        if (n == 2) {            return 2;        }        int nums[] = new int[n + 1];        nums[0] = 1;        nums[1] = 1;        nums[2] = 2;        for (int i = 3; i < nums.length; i++) {            int temp = 0;            for (int j = 0; j < i; j++) {                temp += nums[j] * nums[i - j - 1];            }            nums[i] = temp;        }        return nums[nums.length - 1];    }    public static void main(String[] args) {        // Last executed input: 19        // int numTrees = new UniqueBinarySearchTrees().numTrees(0);// 1        // int numTrees = new UniqueBinarySearchTrees().numTrees(1);// 1        // int numTrees = new UniqueBinarySearchTrees().numTrees(2);// 1        int numTrees = new UniqueBinarySearchTrees().numTrees(3);// 1        // int numTrees = new UniqueBinarySearchTrees().numTrees(6);// 132        // int numTrees = new UniqueBinarySearchTrees().numTrees(19);        System.out.println(numTrees);    }}
0 0